When solving the given system of equations by elimination with addition, which value of k will eliminate the x-terms?2x−y=19k(x+3y)=k(10)

1. Break Down the Problem

We need to manipulate the given system of equations to eliminate the $x$-terms when we add the equations together. The given equations are:

1. $2x - y = 19$
2. $k(x + 3y) = k(10)$

2. Relevant Concepts

To eliminate the $x$-terms, we need to express both equations such that the coefficients of $x$ are opposites. We can rewrite the second equation in standard form first.

3. Analysis and Detail

Let's express the second equation in a more usable form:

• Start with: $k(x + 3y) = k(10)$
• Since $k$ is common, divide both sides by $k$ (assuming $k \neq 0$): $x + 3y = 10$

Now we have the system:

1. $2x - y = 19$
2. $x + 3y = 10$

Next, we need to manipulate these equations. We want the coefficients of $x$ terms to be opposites. To do this, we can multiply the second equation by $-2$:

• Multiply $x + 3y = 10$ by $-2$: $-2x - 6y = -20$

Now, our system looks like:

1. $2x - y = 19$
2. $-2x - 6y = -20$

4. Verify and Summarize

When we add these two equations, the $x$-terms will cancel out: $(2x + (-2x)) + (-y - 6y) = 19 + (-20)$ This simplifies to: $0 - 7y = -1$ Thus, the $x$-terms are eliminated.

For the elimination of $x$ to work, we ensure that the chosen equation for $x + 3y = 10$ is scaled appropriately with respect to the value of $k$.

Final Answer

The value of $k$ can be any non-zero real number since we manipulated the second equation independently. However, to find a specific scenario, we need $k$ such that $k \neq 0$. The derived equations can be manipulated freely as long as $k$ remains constant and non-zero. This implies that $k$ can take any value except 0 for the system to hold.